Solution.

(a) Find the eigenvalues of $A$.

To find the eigenvalues of the matrix $A$, we first compute the characteristic polynomial $p(t)$ of $A$.
We have
\begin{align*}
p(t)&=\det(A-tI)=\begin{vmatrix}
1-t & i\\
-i& 1-t
\end{vmatrix}\\[6pt]
&=(1-t)(1-t)-(i)(-i)=t^2-2t+1-1\\
&=t(t-2).
\end{align*}
Solving $p(t)=0$, we obtain the eigenvalues $0$ and $2$.

(b) For each eigenvalue of $A$, find the eigenvectors.

Let us first find the eigenvectors corresponding to the eigenvalue $0$.
We solve the equation $(A-0I)\mathbf{x}=\mathbf{0}$.
We have
\begin{align*}
A\xrightarrow{R_2+iR_1} \begin{bmatrix}
1 & i\\
0& 0
\end{bmatrix}.
\end{align*}
Hence the eigenvalues are
\[c\begin{bmatrix}
1 \\
i
\end{bmatrix}\]
for any nonzero complex number $c$.

(c) Diagonalize the Hermitian matrix $A$ by a unitary matrix.

To diagonalize the Hermitian matrix $A$ by a unitary matrix $U$, we find an orthonormal basis for each eigenspace of $A$.
As each eigenspace of $A$ is $1$-dimensional by part (b), we just need to normalize any eigenvector for each eigenvalue.

By part (b), we know that $\mathbf{v}_1:=\begin{bmatrix}
1 \\
i
\end{bmatrix}$ is an eigenvector corresponding to the eigenvalue $0$.
The length of this vector is
\[\|\mathbf{v}_1\|=\sqrt{1\cdot 1 +(i)(-i)}=\sqrt{2}.\]
Hence the vector
\[\mathbf{u}_1=\frac{\mathbf{v}_1}{\|\mathbf{v}_1\|}=\frac{1}{\sqrt{2}} \begin{bmatrix}
1 \\
i
\end{bmatrix}\]
is a unit eigenvector.

Similarly, from (b) we see that $\begin{bmatrix}
1 \\
-i
\end{bmatrix}$ is an eigenvector corresponding to the eigenvalue $2$. The length of the vector is $\sqrt{2}$.
Hence
\[\mathbf{u}_2=\frac{1}{\sqrt{2}}\begin{bmatrix}
1 \\
-i
\end{bmatrix}\]
is a unit eigenvector.

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